Interpreting the unit of definite integral I'm calculating a definite integral of a function describing a pressure of certain system where time is the independent variable, that is my function is: 
$$f(t) \;\text{psi},$$
so e.g. $f(0)\;\text{psi} = 5\;\text{psi}$ would mean pressure of $5\;\text{psi}$ at time $t=0$. Now I need to do a definite integral of this function between time interval $[a,b]:$
$$\int^b_a f(t)\;\text{psi}\, dt.$$
Now my question is: What is the unit of this integral? If a client were to ask me how to interpret the result of this integral what would be the correct answer? I tried to think it myself this way:
Since $dt$ is a time quantity I can denote its time unit e.g. by $t$. Then I get my integral to be:   
$$\int^b_a f(t)\;\text{psi}\, dt\, t,$$
so I would read this integral as: "$f(t)$ units of $\text{psi}$ multiplied by $dt$ units of $t$". With this interpretation my integral would be: 
$$\int^b_a f(t)\;\text{psi}\, dt\, t = \left(\int^b_a f(t)\; dt\, \right)\text{psi}\,t = \left[F(b)-F(a)\right]\,\text{psi}\,t,$$
so the end unit would be $\text{psi}\,t$, i.e. pressure times time. If I would divide the end result by the time interval $[a,b]$ I would get the mean $\text{psi}:$ 
$$\frac{F(b)-F(a)}{b-a}\cdot\frac{\text{psi}\,t}{t} = \frac{F(b)-F(a)}{b-a}\cdot \text{psi},$$
so the result would be the average pressure. Is this correct interpretation? 
 A: The interpretation is basically correct, though your notation is rather unconventional and confusing. You're using $t$ both as a variable and as a unit. Since you used a concrete unit, psi, for the pressure, you might as well use a concrete unit like s for the time; if you want to keep the unit general, you should use some letter other than $t$ for denoting it, since you're already using $t$ for the time variable.
There are basically two ways to think about this – the less formal one, which you adopted, is to think of $\mathrm dt$ as a quantity with units of time. The more formal approach would be to regard $\mathrm dt$ as merely a notational device for writing integrals, and to derive the fact that integration over time multiplies the units by units of time from the definition of the integral, e.g. via Riemann sums in which the function values are multiplied by values in units of the independent variable.
The result is the same in either case: in terms of units, integration is like multiplication. Analogously, differentiation is like division, which you can again prove either more formally using the definition if the derivative in terms of limits of difference quotients, or less formally by regarding the differential quotient itself as an actual quotient.
